Sem. 131 Key Solution HW 13 CE 203

Sem. 131 Key Solution HW 13 CE 203

Sem. 131 Key Solution HW 13 CE 203

9–30. The cantilevered rectangular bar is subjected to the force of 20 KN. Determine the principal stress at

points A and B.

30mm 30mm

30mm

30mm

60mm

60mm

300mm

20mm

20mm

20 KN

60mm

120mm

20KN

300mm

N=16KN

Vz=12KN

Mx=3.6KN.m

𝐼 =1

12∗ 60 ∗ 1203 = 8640000 𝑚𝑚4 ,

𝐴 = 120 ∗ 60 = 7200 𝑚𝑚2 ; 𝑄𝐴 = 45 ∗ 30 ∗ 60 = 81000𝑚𝑚3; 𝑄𝐵 = 40 ∗ 40 ∗ 60

= 96000𝑚𝑚3

𝜎𝐴 =𝑃

𝐴+𝑀𝑥 ∗ 𝑧

𝐼=

16 ∗ 1000

7200+

3.6 ∗ 106 ∗ 30

8640000.

𝜎𝐴 = 14.72 𝑀𝑃𝑎.

𝜏𝐴 =𝑉𝑧𝑄𝐴

𝐼𝑡=

12 ∗ 1000 ∗ 81000

8640000 ∗ 60= 1.88 𝑀𝑃𝑎.

𝜎𝑥 = 14.72 𝑀𝑃𝑎; 𝜎𝑦 = 0 ; 𝜏𝑥𝑦 = 1.88 𝑀𝑃𝑎

𝜎1,2 =𝜎𝑥 + 𝜎𝑦

2±

𝜎𝑥 − 𝜎𝑦

2

2

+ 𝜏𝑥𝑦2

𝜎1,2 =14.72 + 0

2±

14.72 − 0

2

2

+ 1.882 = 7.36 ± 7.60

𝝈𝟏 = 𝟏𝟒.𝟗𝟔 𝑴𝑷𝒂 𝑨𝒏𝒔. ; 𝝈𝟐 = −𝟎.𝟐𝟒 𝑴𝑷𝒂 𝑨𝒏𝒔

𝜎𝐵 =𝑃

𝐴−𝑀𝑥 ∗ 𝑧

𝐼=

16 ∗ 1000

7200+

3.6 ∗ 106 ∗ 20

8640000.

𝜎𝐴 = −6.11 𝑀𝑃𝑎.

𝜏𝐴 =𝑉𝑧𝑄𝐴

𝐼𝑡=

12 ∗ 1000 ∗ 96000

8640000 ∗ 60= 2.22 𝑀𝑃𝑎.

𝜎𝑥 = −6.11 𝑀𝑃𝑎; 𝜎𝑦 = 0 ; 𝜏𝑥𝑦 = 2.22 𝑀𝑃𝑎

𝜎1,2 =𝜎𝑥 + 𝜎𝑦

2±

𝜎𝑥 − 𝜎𝑦

2

2

+ 𝜏𝑥𝑦2

𝜎1,2 =−6.11 + 0

2±

−6.11 − 0

2

2

+ 2.222

𝝈𝟏 = 𝟎.𝟕𝟐 𝑴𝑷𝒂 𝑨𝒏𝒔. ; 𝝈𝟐 = −𝟔.𝟖𝟑 𝑴𝑷𝒂 𝑨𝒏𝒔

Point A.

Point B.

Sem. 131 Key Solution HW 13 CE 203

Sem. 131 Key Solution HW 13 CE 203

9–68. Draw Mohr’s circle that describes each of the following states of stress.

kPa

kPa

MPa

kPa 700 kPa 600 kPa

650 kPa

50 kPa

4 MPa

2 MPa

2 MPa

Sem. 131 Key Solution HW 13 CE 203

(kPa)

(kPa)

(MPa)

(MPa)

Sem. 131 Key Solution HW 13 CE 203

9–75. The 50 mm.-diameter drive shaft AB on the helicopter is subjected to an axial tension of 50 KN and a

torque of 0.45 KN.m Determine the principal stress and the maximum in-plane shear stress that act at a

point on the surface of the shaft.

𝜎 =𝑃

𝐴= 50 ∗

1000

𝜋 ∗ 252= 25.47 𝑀𝑃𝑎.

𝜏 =𝑇𝑐

𝐽=

0.45 ∗ 106 ∗ 25𝜋2 ∗ 254

= 18.34 𝑀𝑃𝑎.

𝜎1,2 =𝜎𝑥 + 𝜎𝑦

2±

𝜎𝑥 − 𝜎𝑦

2

2

+ 𝜏𝑥𝑦2

𝜎1,2 =25.47 + 0

2±

25.47 − 0

2

2

+ 18.342 = 12.74 ± 22.33

𝝈𝟏 = 𝟑𝟓.𝟎𝟕 𝑴𝑷𝒂 𝑨𝒏𝒔. ; 𝝈𝟐 = −𝟗.𝟓𝟗 𝑴𝑷𝒂 𝑨𝒏𝒔.

𝜏max 𝑖𝑛−𝑝𝑙𝑎𝑛𝑒 = 𝜎𝑥 − 𝜎𝑦

2

2

+ 𝜏𝑥𝑦2 =

25.47 − 0

2

2

+ 18.342

= 𝟐𝟐.𝟑𝟑 𝑴𝑷𝒂. 𝑨𝒏𝒔.